Functions #4 - Questions and Answers

Question 1 What number exceeds its square by the largest amount? [Hint: Find the number x that maximizes f(x) = x - x².] Answer Let the number is x, then f(x) = x - x² The goal is to find the absolute maximum of this function. Since f'(x) = 1 - 2x Then f'(x) = 0 for x = 0.5 and f"(x) = -2 => f"(0.5)<0 Hence an absolute maximum point occurs for the given value of x and f(0.5) = 0.25 Question 2 Determine the set of points at which the function is continuous. f(x,y) = (x,y) not equal to (0,0) and 1 if (x,y)=(0,0) Answer 1 x²y³ / (2x² + y²) On R³ without the origin, f is a rational function, and so is continuous at every point in its domain. Since it is defined on this entire domain, we see that f is continuous at every point. To check whether f is continuous at the origin, we must check whether lim(x,y)→(0,0) f(x, y) = f(0,0). Checking along the line y=0, we see that limx→0 x²(0)³/(2x²+(0)²) = limx→0 0 = 0 But this implies that if f has a limit as (x,y) tends to (0,0), it must be 0, which is not the value of f(0,0). Thus we see that in this case, f(x,y) is not continuous at the origin. Result: f(x,y) is not continuous at the origin. Answer 2 Explanation: To determine the set of points at which the function is continuous, we need to examine the behavior of the function at different points. The given function is: f(x,y) = { x²y³ / (2x² + y²) if (x, y) ≠ (0,0) 1, if (x, y) = (0, 0) Let's first analyze the behavior of the function when (x,y) ≠ (0,0). In this case, the function is defined as: f(x,y) = x²y³ / (2x² + y²) We notice that the function is a rational function, which is continuous everywhere except for the points where the denominator is equal to zero. Hence, we need to determine when 2x² + y² = 0. Since 2x² is always non-negative, for the sum 2x² + y² to be zero, both 2x² and y² must be zero. This implies x = 0 and y = 0. However, this contradicts the assumption that (x,y) ≠ (0,0). Therefore, for all points (x,y) ≠ (0,0), the function f(x, y) is continuous. Now, let's analyze the behavior of the function at the point (0,0). In this case, the function is defined as: f(x,y) = 1 Since the function is defined as a constant, it is continuous at (0, 0). In summary, the function f(x, y) is continuous at all points except (0,0). The set of points at which the function is continuous is given by: {(x,y) ∈ R²:(x, y) ≠ (0,0)} Answer 3 Step 1: To determine the set of points at which the function is continuous, let's analyze the function f(x,y) = x²y³ / (2x² + y²) for (x, y) ≠ (0,0) and 1 for (x,y) = (0,0). First, we need to check the continuity of the function at (0,0). To do this, we'll examine the limit of the function as (x, y) approaches (0,0). lim(x,y)→(0,0) f(x,y) = lim(x,y)→(0,0) x²y³/(2x²+y²) To evaluate this limit, we can use polar coordinates. Let's substitute x = rcosθ and y = rsinθ into the function: limr→0 (rcosθ)² (rsinθ)³/(2(rcosθ)² + (rsinθ)²) Simplifying this expression gives: limr→0 r⁵cos²θsin³θ/(2r²cos²θ + r²sin²θ) Now, we can cancel out the r² term from the denominator: limr→0 r³cos²θsin³θ/(2cos²θ + sin²θ) Step 2: Next, we can take the limit as r approaches 0: limr→0 r³cos²θsin³θ = 0 Thus, the limit of the function as (x,y) approaches (0,0) exists and is equal to 0. Therefore, the function is continuous at (0,0). Now, let's consider the continuity of the function for (x, y) ≠ (0,0). Since the function is a rational function, it is continuous everywhere except where the denominator is equal to 0. Setting the denominator 2x² + y² equal to 0, we have: 2x² + y² = 0 Since both x² and y² are non-negative, the equation above cannot hold for any real values of x and y. Thus, the denominator is never equal to 0 for (x, y) ≠ (0,0). Therefore, the function is continuous for all (x, y) ∈ R². Answer 4 To determine the set of points at which the function is continuous, we need to examine the behavior of the function as we approach different points. Let's analyze the function f(x,y) = x²y³ / (2x² + y²) when (x, y) is not equal to (0,0), and f(x,y) = 1 when (x, y) = (0,0). First, let's investigate the function's behavior as (x,y) approaches (0,0). We'll consider approaching along different paths. Approaching along the x-axis, we let y = 0. Substituting into the function, we have: limx→0 f(x, 0) = limx→0 x².0³ / (2x²+0²) = limx→0 0 / 2x² = 0 Approaching along the y-axis, we let x = 0. Substituting into the function, we have: limy→0 f(0,y) = limy→0 0².y³ / (2.0² + y²) = limy→0 0 / y² = 0 Thus, the function approaches 0 as we approach (0,0) along the x-axis and the y-axis. To determine if the function is continuous at (0,0), we need to check if the limit of the function as (x, y) approaches (0, 0) exists and is equal to the value of the function at (0,0). Let's compute the limit of the function as (x, y) approaches (0,0) along the path y = mx, where m is a constant: lim(x,y)→(0,0) f(x,y) = limx→0 x²(mx)³ / (2x² + (mx)²) = limx→0 m³x⁵ / (2x² + m²x²) = limx→0 m³x³ / (2 + m²) = (m³*0)/(2 + m²) = 0 Since the limit is 0 for all paths, the limit exists and is equal to the value of the function at (0,0), which is 1. Therefore, f(x, y) is continuous at (0, 0). In conclusion, the function f(x,y) = x²y³ / (2x² + y²) is continuous for all points (x,y) in the entire xy- plane, including (0,0). Question 3 Find the indicated difference quotient and simplify your answer. f(x) = x³ + 3x , (f(x + h) – f(x))/h, h does not equal 0 f(x) = 1/x², (g(x) - g(3))/(x-3), x does not equal 3. f(t) = 1/(t - 2) , (f(t) - f(1))/(t-1), t does not equal 1. Answer 1 Using given f(x) first find f(x+h). f(x) = x³ + 3x f(x+h) = (x+h)³ + 3(x+h) f(x+h) = x³ + 3x²h + 3xh² + h³ + 3x + 3h Plugged the expression of f(x) and f(x+h) in the difference quotient and simplified. (f(x+h) - f(x))/h = (x³ + 3x²h + 3xh² + h³ + 3x + 3h - (x³ + 3x))/h = (3x²h + 3xh² + h³ + 3h)/h = h(3x² + 3xh + h² + 3)/h = 3x² + 3xh + h² + 3 Answer: 3x² + 3xh + h² + 3 Answer 2 Using given g(x) first find g(3) g(x) = 1/x² g(3) = 1/3² g(3) = 1/9 Plugged the expression of g(x) and g(3) in the difference quotient and simplified. (g(x) - g(3))/(x-3) = (1/x² - 1/9)/(x-3) = (9 - x²)/(9x²(x - 3)) = -(x² - 9)/(9x²(x-3)) = -(x + 3)(x - 3) / (9x²(x - 3)) = -(x+3)/(9x²) Answer: -(x+3)/(9x²) Question 4 How do you evaluate g(-5) given g(x) = 3x? Answer 1 Explanation: to evaluate g(-5) substitute x = -5 into g(x) => g(-5) = 3 * -5 = -15 Answer 2 Given: x = -5 than g(-5) = 3 * -5 g = -15 g = 3 ans Question 5 What is the slope of a line perpendicular to the line whose equation is x - y = 5. Answer Slopes on perpendicular lines are reciprocally negative. First, by finding y, we write the given equation in slope intercept form: x - y = 5 x - 5 = y or y = x - 5 So, the slope of the given line is 1 which means that the slope of the line perpendicular to it is: -(1/1) = -1 22.9s